MATH 4 (CALCULUS II) IN ORDER TO OBTAIN A PERFECT SCORE IN ANDROULAKIS MATH 4 CLASS YOU NEED TO MEMORIZE THIS HANDOUT AND SOLVE THE ASSIGNED HOMEWORK ON YOUR OWN GEORGE ANDROULAKIS TRIGONOMETRY θ sin(θ) cos(θ) tan(θ) 0 0 0 π 6 π 4 π 3 π 3 3 3 3 3 0 - Pythagorean Theorem: sin (θ) + cos (θ) =, + tan (θ) = sec (θ), + cot (θ) = csc (θ). The sine of the double angle: sin(θ) = sin(θ) cos(θ) The square of the sine or the cosine: cos (θ) = +cos(θ), sin (θ) = cos(θ). THE 7 INDETERMINATE FORMS OF LIMITS (), 0 0 : usually we use L Hospital s rule. Two important such limits are lim cos x lim = x 0 x. () 0 : we turn the product into a quotient by fg = f g = g f x 0 sin x x = and and we reduce the problem to (). (3) : usually we add the fractions by creating a common denominator, or we multiply and divide by the conjugate expression if we have the difference of two square roots, and usually the problem is reduced to (). (4), 0 0, 0 : we call y the function that we try to compute its limit, we consider ln y which turns the power into a product of the form 0 and we continue as in (). At the end we exponentiate the answer. One important limit of the form is lim x 0 ( + x )x = e.
DERIVATIVES AND INTEGRALS OF COMMONLY USED FUNCTIONS F f f x n+ n+ + C if n ; ln x + C if n = xn nx n a x ln a + C ax, 0 < a a x ln a e x + C e x e x ln x ln a x ln a NNTM log a x = NNTM ln x x cos x sin x cos x sin x cos x sin x NNTM tan x sec x NNTM cot x csc x ln sec x + tan x + C sec x sec x tan x ln csc x + cot x + C = ln csc x cot x + C csc x csc x cot x NNTM arcsin x x NNTM arctan x +x NNTM arcsec x x x NNTM stands for Not Need To Memorize Remark 0.. Observe that the derivative makes the powers (i.e. the functions of the form x n ) better (since (x n ) has a smaller power). Also the derivative makes the logarithmic functions and the inverse trigonometric functions much better (this is a philosophical comment but you can notice that you can compute their derivatives for more points that we can compute the functions themselves without the use of calculators). This can be used as a mnemonics for part of the above table, but also it will be useful for Integration By Parts when you will try to decide which function to differentiate and which to integrate. INTEGRATION BY SUBSTITUTION f(g(x))g(x) dx = f(u)du where u = g(x) and b a f(g(x))g(x) dx = g(b) g(a) f(u)du. Remark 0.. It is very important when you set u = g(x) that you see not only the u inside the integral, but you also see the du (i.e. you see the g(x) dx. If you do not see this part, then the integration by substitution method will not work!
Topic : INTEGRATION BY PARTS (IBP) Int. f(x) g(x) Diff. dx = F (x)g(x) F (x)g(x) dx and b a g(x) dx = F (x)g(x) Int. f(x) Diff. b a b a F (x)g(x) dx. Remark 0.3. Note that I always write on top of each function Diff. or Int. That helps me to remember which function to integrate and which to differentiate. I strongly recommend that you do the same! Remark 0.4. If you do IBP twice then you should not change your mind in the middle of the way on which function to integrate and which function to differentiate because you will end up with nothing as in: Int. Diff. Diff. Int. f(x) g(x) dx = F (x)g(x) F (x) g(x) dx ( = F (x)g(x) F (x)g(x) = f(x)g(x)dx. ) f(x)g(x)dx 3
Remark 0.5. The shortcut for performing iterated IBPs one after the other is illustrated by the following table, (here F is the integral of f; F is the integral of F ; F is the integral of F; I run out of ideas on how to denote integrals after that) Int. Diff. f(x) g(x) + F (x) g(x) F(x) g(x) + F(x) g(x) The above table gives: Int. Diff. f(x) g(x) dx = +F (x)g(x) F(x)g(x) + F(x)g(x) F(x)g(x) dx. That is 3 IBPs. Notice that the products without integrals correspond to slanted products where it is slanted down on the side of the integration (and slanted up on the side of differentiation). Also notice the column of the alternating signs. Finally notice that the last horizontal (not slanted) row of the table corresponds to an integral. We can do this iteration (i.e. multiple IBPs) as many times as we want! For example, if g(x) is a power of x (e.g. g(x) = x 4 ), then if we perform 5 consecutive IBPs then g(x) becomes 0 so the last integral, (which corresponds to the last horizontal line of the table), is equal to 0 (so we only obtain the 5 slanted products and not an integral since the integral corresponding to the last horizontal line of the table becomes 0). Sometimes, when we perform consecutive IBP s it is helpful to shift terms from one side to the other between the columns labeled Int. and Diff.. This can be seen for example if you compute x ln 3 xdx. You will need to perform 3 consecutive IBP s. Everytime, the derivative of the logarithmic function will produce a term x which should be moved to join the function which is integrated. You should perform these consecutive IBP s! 4
Topic : INTEGRATION OF TRIGONOMETRIC FUNCTIONS The most commonly used trick here is the method of u-substistution. Integrals of expressions of sines and cosines: Try the substitution u = sin x or u = cos x, or try using the double angle formulas cos x = + cos(x), sin x = cos(x). It will be useful to remember the Pythagorean Theorem sin x + cos x = which helps you to convert between even powers of sines and cosines. Integrals of expressions of tangents and secants: Try the substitution u = tan x or u = sec x, or try converting into an expression of sines and cosines. It will be useful to remember the Pythagorean Theorem + tan x = sec x which helps you to convert between even powers of tangents and secants. This will be handy for example in integrals like tan n xdx where you first want to create some secants which are needed for the u- substitution but they are absent in this integral. Integrals of expressions of cotangents and cosecants: Try the substitution u = cot x or u = csc x, or try converting into an expression of sines and cosines. It will be useful to remember the Pythagorean Theorem +cot x = csc x which helps you to convert between even powers of cotangents and cosecants. The integrals sec odd x tan even xdx (for example sec 3 xdx, sec x tan xdx, sec x tan 4 xdx, sec 3 x tan 4 xdx) needs its own trick: First IBP. Then use the Pythagorean formula + tan x = sec x. And finally use algebra to solve for the integral that you had to compute. Warning: The higher the total power is, the more times you have to repeat this process! (So the easiest of those integrals are the sec 3 xdx and sec x tan xdx where the total power is equal to 3). 5
Topic 3: INTEGRATION OF EXPRESSIONS OF ax + bx + c Here we use the method of trigonometric substitution which is described now: If b 0 then first complete the square in order to get rid of the term bx. completion of the square: Recall the ( ax + bx + c = a ( = a x + b a x + c a x + b ( ( = a x + b a = a ) ( b a x + a ) b ( ( x + b ) + a ) ( b a ) 4a + c a ) 4ac b 4a. ) ) + c a If b = 0 then the integrant has one of the following three forms: () ax + c with a > 0 and c > 0. Then set ax = c tan θ and use the identity + tan θ = sec θ. () ax c with a > 0 and c > 0. Then set ax = c sec θ and use the identity sec θ = tan θ. (3) ax + c with a > 0 and c > 0. Then set ax = c sin θ and use the identity sin θ = cos θ. 6
Topic 4: INTEGRATION OF RATIONAL FUNCTIONS First make sure that the rational function is proper, i.e. the degree of the numerator is smaller than the degree of the denominator. If this is not the case, i.e. if the degree of the numerator is larger than or equal to the degree of the denominator then first perform long division. Then rewrite the integrant as numerator denominator = quotient + reminder denominator. reminder denominator Then the rational function is proper, and next we show how to integrate that function. Let p(x) q(x) be a proper rational function. A theorem, called Fundamental Theorem of Algebra, says that it is always possible to factor the denominator as in the example: q(x) = (x R ) 3 (x R )(x + x + ) (x + 5). Here R and R roots of p(x), (i.e. q(r ) = q(r ) = 0), 3 is the multiplicity of the root R, and is the multiplicity of the root R. The quadratics x + and x +5 are irreducible, i.e. they do not have real roots, since their discriminants are negative. In general to factor a polynomial of degree at least equal to 3 is very difficult. A theorem says that if you have a polynomial with integer coefficients, then possible rational roots are among the numbers that you get by dividing the possible integer divisors of the constant term, by the possible integer divisors of the coefficient of the highest term; then a synthetic division, instead of the long division, can be used to find out which of these possible roots are actually roots. For example, in order to factor 6x 3 + x 5x, notice that the constant term is and its integer divisors are ±, ±. Notice also that the coefficient of its highest term is 6 and its integer divisors are ±, ±, ±3, ±6. If you divide the integer divisors of the constant by the integer divisors of the coefficient 6 of the highest term you obtain the set of numbers {±, ±, ±, ± 3, ± 6, ± 3 } (that has numbers). If you try synthetic division you can see that that the numbers, and 3 are roots of 6x 3 + x 5x while the other 9 numbers are not roots. Thus the polynomial 6x 3 + x 5x can be factored as 6(x )(x + )(x + 3 ) = (x )(x + )3(x + ) = (x )(x + )(3x + ). 3 Another trick for factoring polynomials is the trick of grouping terms. For example if you want to factor x 3 x + x then you group the first two terms and you obtain x (x ), and similarly from the last two terms you obtain (x ). If you put these together you obtain x 3 x + x = x (x ) + (x ) = (x + )(x ). In class we will not relay on any of these two methods. Instead, I would like the students to know the identities: and x n a n = (x a)(x n + x n a + x n 3 a + + xa n + a n ), x n+ + a n+ = (x + a)(x n x n a + x n a xa n + a n ), which are valid for any positive integer n, and any real numbers x and a. Some applications of these formulas are the following: x 4 = (x )(x 3 + x + x + ), x 4 6 = (x )(x 3 + x + 4x + 8), x 3 + 7 = (x + 3)(x 3x + 9). 7
Once the rational function is proper and the denominator has been factored as above, the method of partial fractions guarantees that the rational function can break into sums of partial fractions such that the denominator of each partial fraction is either of the form A Ax+B (Bx+C) or of the form p (Cx +Dx+F ) p where Cx + Dx + F is an irreducible quadratic (i.e. its descriminant D 4CF is negative and therefore it does not have real roots and cannot be factored without using complex numbers). For instance, when q(x) = (x R ) 3 (x R )(x + x + ) (x + 5). there exist constants A, B, C, D, E, F, G, H, I such that for every real number x, (except R and R ), we have: p(x) q(x) = A x R + B (x R ) + C (x R ) 3 + Dx + E x + x + + F x + G (x + x + ) + Hx + I x + 5. In order to compute the unknown constants A,..., I, first we multiply the above equation by q(x) to get rid of all denominators. Since we have 9 unknowns (A,..., I), we need 9 equations. We plug in x = R and x = R and we obtain equations. Then we plug in 7 more values of x to obtain the total of 9 equations that we need. We solve these 9 equations and we compute A,..., I. Then we integrate each of the above 6 partial fractions that were produced. The first 3 partial fractions are easy to integrate. Dx+E The fourth partial fraction, x +x+ is easy to integrate if Dx+E = (x +x+) = x+. Indeed, if Dx + E = x + then Dx + E x + x + = x + x + x + = du u = ln u + C = ln x + x + + C = ln(x + x + ) + C Dx+E x +x+ by setting u = x + x +. Also it is easy to integrate if D = 0 since then you complete the square x + x + = (x + ) + 3 4 and you use that du u + a = ( u ) a arctan + C, a to obtain ( ) dx x + x + x + = arctan + C. 3 In general, Dx + E may not be equal to neither x + nor E, but we can use some algebra to write Dx + E x + x + dx = D x + x + x + dx + (E D ) x + x + dx ) = D ln(x + x + ) + (E D ) arctan ( x + 3 + C. You should remember the above trick: If you have a polynomial of first degree in the numerator and a power of an irreducible quadratic in the denominator, then you can always break the fraction into two fractions: the first will have the exact derivative of the irreducible quadratic on the numerator, and the second will have a constant on the numerator. In order to compute the integral of the first you use a u substitution and you set u to be equal to the irreducible quadratic. In order to compute the second integral, you complete the square in the irreducible quadratic and you use the tangent trigonometric substitution. 8
F x+g The same trick can be applied to the fifth partial fraction to break it, followed (x +x+) by the completion of the square for x + x + and the trig substitution x + = 3 tan θ: F x + G (x + x + ) dx = F x + (x + x + ) dx + (G F ) = F (x + x + ) + (G F ) (x + x + ) dx ((x + ) + 3 4 = F (x + x + ) + (G F 3 ) sec θdθ ( 3 4 tan θ + 3 4 ) = F (x + x + ) + (G F 3 ) ( 3 4 ) = F (x + x + ) + (G F )4 3 3 cos θdθ ) dx + cos(θ) dθ = F (x + x + ) + (G F )4 3 3 (θ + sin(θ) ) + C ( ) 4 = x + arctan 3 x + + 3 4 (x + x + ) + C. The integral of the sixth partial fraction Hx+I x +5 of the fourth partial fraction. can be computed similarly to the integral 9
SERIES Limits of sequences A sequence (a n ) n N is just a function defined on the positive integers N, (i.e. the domain of a sequence function is equal to N). If (a n ) n N is a sequence then in order to compute the limit lim n a n of the sequence (a n ) n N, then most of the time we replace the positive integer n by a positive real x so the sequence (a n ) n N is transformed into a function f(x). Then instead of computing the lim n a n we compute lim x f(x). Sometimes, instead of replacing n by x, we set x to be a more complicated expression of n which appears more than once in the formula of the sequence, (e.g. we replace /n with x and then we consider a limit of a function as x 0). Let n= a n be a series, and let S N = N n= a n be its partial sums for every N N. Q: What does it mean that a series converge? A: The series n= a n converges means that lim N S N number L. Then we write n= a n = L. exists and lim N S N is a real Q: What does it mean that a series diverge? A: The series n= a n diverges if either the limit lim N S N does not exist, (for example if a n = ( ) n then the partial sums S N oscillate wildly between and 0 so the partial sums do not converge), or the limit lim N S N exists but it is equal to +, (for example if a n = for all n then S N = N ), or equal to, (for example if a n = for all n then S N = N ). Note that if a n 0 for all n then (S N ) N N is increasing because S N+ = S N + a N+ S N + 0 = S N, so lim N S N will always exist (but it may be equal to + ). Thus if a n 0 for every n, then n= a n converges means that lim N S N <. That is why, when a n 0 for all n, then instead of writing the series n= a n converges sometimes we equivalently write n= a n <. We never replace the phrase the series n= a n converges by the phrase n= a n < when some a n s are negative! 0
First recall the Geometric Series. Perhaps this is the simplest examples of series because we can actually compute its partial sums! Geometric series The partial sums S N = N an of the geometric series an are given by the formula N = an+ a, and therefore, the geometric series n= an converges if a (, ), or diverge if a (, ] [, ). Another type of series for which we can compute their partial sums are the telescopic series.
FLASH CARDS FOR TESTS OF CONVERGENCE AND DIVERGENCE Memorize verbatim those cards and fill the blanks appropriately in every problem of the exams. Assume that you are asked: Does the series n= a n converge or diverge? The next flash cards tell you how you should answer this question. Topics 5-7: n th -TERM TEST, or DIVERGENCE CRITERION (DC) Your Answer: Let a n =. Then lim n a n = 0. Therefore by the DC, the series n a n diverges. Note that the second box in the DC, is allowed to be replaced by DNE (Does Not Exist). If the limit exists then you should compute it. Some useful limits to remember are: sin x lim x 0 x cos x =, lim = x 0 x, lim ( + x x )x = e.
Topic 5: INTEGRAL TEST (IT) Your Answer: Let f : [, ) R be defined by f(x) =. Then f is continuous, positive, decreasing and f(n) = a n for all n N. f(x)dx =. Thus f(x)dx =. Therefore by the integral test, the series. To figure out the formula of the function f(x), (i.e the content of the second box), we change the n in the formula of a n into an x. The content of the first box is not necessarily n 0 ; it can be larger than n 0 and has to be chosen in such way that the function f is continuous, positive and decreasing on that domain. In the exams you will not have to prove that the function f is continuous, positive, decreasing, and satisfies f(n) = a n for all n N, but you have to check it (and write the correct domain on which this happens). The content of the last box will be converges if in the third box from the end you find a finite number. The content of the last box will be diverges if in the third box from the end you find. A consequence of the IT is the p-series test: p-series test If p > then n= If p then n= n p n p converges. diverges. 3
Topic 6: COMPARISON TEST (CT) (Version ) Your Answer: 0 a n b n = for all n N and n= b n converges because. Therefore by the CT, the series n= a n converges. Topic 6: COMPARISON TEST (CT) (Version ) Your Answer: 0 b n = a n for all n N and n= b n diverges because. Therefore by the CT, the series n= a n diverges. Topic 6: LIMIT COMPARISON TEST (LCT) Your Answer: We have that a n 0. Let b n = 0. a n lim = which is 0 and. n b n We have that n= b n because. Thus by the LCT, n= a n. Sometimes the following limits are useful for the LCT: sin x lim x 0 x, lim cos x = x 0 x. Most of the time the series n= b n that we pick in the CT or LCT is a p-series or a geometric series, but also one should not exclude series that can be handled using other tests such as the IT! When you use the CT and the sequence (a n ) contains at least two of logarithms, powers, exponential, then remember that any positive power of ln n is eventually less than than any positive power of n which is eventually less that any exponential, which is eventually less than n! which is less than n n, (where I assume that the exponential has base > ). For example (ln n),000 n for large enough n, (e.g. n 5 0 8,59 works). Also ln(n,000 ) n for large enough n because ln(n,000 ) =, 000 ln n which is n for n 4 0 8. Also n,000 n for large enough n, (e.g. n 3, 747 works here). 4
For the next flashcard, assume that you are asked: Does the series n= (( )n a n converge or diverge? Topic 7: ALTERNATING SERIES TEST (AST) Your Answer: a n = converges. 0 therefore by the AST the series n= ( )n a n 5
Topic 7: ABSOLUTE CONVERGENCE, CONDITIONAL CONVERGENCE, DIVERGENCE Note that you can apply the AST to prove that the series n= converges, even thought by the p-series test, the series n= diverges. Thus you can think of the part n ( )n as a helping factor that helps the series n= to converge. That is indeed the case because n if ( ) n is present then the AST may be applicable! Thus we started with a series n= which does not converge and after we applied the helping factor ( ) n, we obtained a new series which converges. You can also think of the previous example in the opposite way: You start from the series ( ) n n= which converges (as it can easily be seen by the AST), and apply the absolute n value inside the series to obtain ( ) n n n ( ) n n = n n= n= which diverges. Thus the addition of the absolute value inside the series ( ) n n= prevents n the series from converging any more. There is a Theorem (i.e. a statement that can be proved) that says that every time that we add absolute values inside a series we make it more difficult (or even impossible!) for the series to converge. So there are three types of series: Series that converge even if we add absolute values inside the series. These series are called absolutely convergent series Series that converge as they are, but if you add absolute values inside the series, then they do not converge any more. We can think of these series as being fragile because the addition of the absolute values destroys their convergence. These series are called conditionally convergent series. Series that do not converge. These series are called divergent. What the above list does, is that it breaks the previously known group of converging series into two groups: the absolutely convergent series and the conditionally convergent series. Note that if a n 0 for all n, then a n = a n hence the addition of absolute values inside the series does not change the series. Thus, for those series, converging and absolutely converging are the same notions, and the conditionally convergent scenario never happens! Hence if the series has only positive terms then there are only two possible scenarios for the behavior of the series: convergent or divergent; but if the series has some positive and some negative terms then there are three possible scenarios for the behavior of the series: absolute convergent, conditional convergent or divergent. 6
For the next flashcard, assume that you are asked: Does the series n= (( )n a n converge (absolutely or conditionally) or diverge? Topic 7: RATIO TEST (RaT) (Version ) Your Answer: lim n a n+ a n = <. Thus by the Ratio test the series n= a n converges absolutely. Topic 7: RATIO TEST (RaT) (Version ) Your Answer: lim n a n+ a n = >. Thus by the Ratio test the series n= a n diverges. Topic 7: ROOT TEST (RoT) (Version ) Your Answer: lim n n an = <. Thus by the Root test the series n= a n converges absolutely. Topic 7: ROOT TEST (RoT) (Version ) Your Answer: lim n n an = >. Thus by the Root test the series n= a n diverges. Usually we use the Ratio test when the formula of the terms of the series contain factorials. Also usually we use the Root test when a n has the form a n = ( ) n. Then n a n = n ( )n =. 7
Methodology for determining if a series converges absolutely, conditionally or diverges. For the question: Does the series n= a n converge absolutely, conditionally, or diverge, I recommend that you follow the next steps in the order described: Step : Spend the first 30 seconds to examine if lim n a n 0. If this happens then by the n th -term test the series n= a n diverges and you are done. If this does not happen proceed to the next step. Step : Spend the next 30 seconds to examine if the Ratio or Root Tests are applicable. If they are, then the series n= a n will be found to converge absolutely (if version of these tests was applicable) or to diverge (if version of these tests was applicable). If this happens you are done. If these tests are not applicable proceed to the next step. Step 3: Examine whether the series n= a n converges or diverges using either of the tests: IT, p-series test, CT, or LCT. If you find that n= a n converges then n= a n converges absolutely and you are done. If n= a n diverges then hopefully the AST will be applicable and will give you that n= a n converges and therefore n= a n converges conditionally. Remark: It is important to notice that in the suggested methodology I tried the Ratio or Root test before trying the IT, p-series test, CT, or LCT. The reason is that the Version of the Ratio and Root test instead of giving that n= a n diverges, it actually gives that n= a n diverges (which is a stronger conclusion than n= a n diverges! Thus, in Step when you apply the Version of Ratio or Root test then you obtain that n= a n diverges, and that finishes the problem! But if in Step when had instead applied the p-series test, or the IT, or CT or LCT then you could have found that n= a n diverges which does not finish the problem since it does not tell you that n= a n diverges. Remark: I did not mention the geometric series test in the methodology above, because every geometric series can be examined using the Ratio and Root tests. So if you insist to include the geometric series test in the above methodology then you include it in Step. If you include any of the following phrases in your answer then your whole answer of that question will receive a zero! By AST the series diverges. By AST the series converges absolutely. By RaT the series converges conditionally. By RoT the series converges conditionally. 8
Topic 8: INTERVAL OF CONVERGENCE OF A TAYLOR SERIES In order to compute the interval of convergence of a Taylor series a n (x x 0 ) n, you use the Ratio test. This will give you the radius of convergence R such that the Taylor series will converge absolutely in the interval (x 0 R, x 0 +R). Then you test the end points x = x 0 R and x = x 0 + R by using any of the tests for convergence or divergence of series (different than the Ratio and Root test). 9
Topic 9: WRITE THE TAYLOR SERIES OF A FUNCTION (sections 0.7 and 0.9) Question: Given a function f and a center x 0, find coefficients (a n ) n N such that f(x) = a n(x x 0 ) n for all x s in an interval around x 0 (the interval of convergence of the Taylor series). The methodology for answering this question, which is given to us in section 0.7 is the following: We start from x = x n for < x < which follows from the formula for the sum of the geometric series. This is the Taylor series of the function with center 0. If x x 0 = 0 then we manipulate the function and x simultaneously we manipulate similarly its power series, until we arrive at the function f(x). All the possible manipulations are: Replace x by a power of x. Multiply by a power of x. Differentiate. When you differentiate you may loose one or two points from the interval of convergence. Integrate. When you integrate you may end up with a larger interval of convergence (by one or two points). If x 0 0 then we first set x x 0 = z, i.e. x = z + x 0 and the new variable becomes z. We perform manipulations as above (where you now have z instead of x) until you arrive at the function f(z). Finally you replace z by x x 0. Examples: () Replace x by x in the power series of x to obtain + x = ( ) n x n for < x <. This is the Taylor series of the function with center x +x 0 = 0. () Square the power series of or (even easier!) differentiate it, to obtain x ( x) = nx n for < x <. n= This is the Taylor series of the function ( x) with center x 0 = 0. (3) Replace x by x in () to obtain + x = ( ) n x n for < x <. This is the Taylor series of the function with center x +x 0 = 0. (4) Integrate (3) to obtain arctan x = ( ) n xn+ n + 0 + C for < x <.
By pluging x = 0 obtain that C = 0 i.e. arctan x = ( ) n xn+ n + for < x <. Observe that the last formula is even valid for x = because the series on the right hand side converges by AST. Hence the last formula is valid for < x which gives the Taylor series of the function arctan x with center x 0 = 0. Further practice: Compute the Taylor series of the functions with center x 0 = 0: x, x, 3 ( x), x ln( + x), ln( x), ln( + x), 3 x x. Can you write the above examples with center x 0 = 3? Can you find the Taylor series of +x with center x x 0 = 0? An alternative way to compute the Taylor series of a function f with a given center x 0 is given in section 0.9: The Taylor series of the function f with a center x 0 is given by a n (x x 0 ) n where a n = f (n) (x 0 ) n! for n = 0,,.... Please memorize the following Taylor series: x = x n = + x + x + x 3 +, < x < e x = x n n! = + x + x + x3 3! +, x R sin x = ( ) n (n + )! xn+ = x x3 3! + x5 5! x7 7! +, x R cos x = ( ) n (n)! xn = x! + x4 4! x6 6! +, x R. You can only start from one of these to produce others in the exams.
Topic 0: ERROR IN APPROXIMATION OF A FUNCTION BY TAYLOR POLYNOMIALS Usually we write p N (x) = N f (n) (x 0 ) (x x 0 ) n n! to denote the Nth Taylor polynomial of a function f. Of course in order this polynomial to make sense, the function f has to have at least N many derivatives (or, f is differentiable at least N many times). If the function f is differentiable at least N + many times, then f(x) p N (x) = f (N+) (c) (N + )! (x x 0) N+ for some c between x 0 and x. Sometimes we denote R N (x) = f (N+) (c) (N + )! (x x 0) N+, and we call R N (x) the remainder of f when f is approxiamated by the Nth Taylor polynomial. Notice that the number c in the remainder depends on the number x. Thus for fixed x, if f is a function which has infinitely many derivatives and max{ f (N+) (y) : for all y between x 0 and x} (x x 0 ) N+ 0 as N, (N + )! then the Taylor series In particular, for fixed R > 0, if then the Taylor series (x 0 R, x 0 + R). f (n) (x 0 ) n! (x x 0 ) n of the function f converges to f(x). max{ f (N+) (y) : x 0 R < y < x 0 + R} R N+ 0 as N. (N + )! f (n) (x 0 ) n! (x x 0 ) n of the function f converges to f(x) for all x As a special case, if there exists R > 0 and a finite number M such that max{ f (N+) (y) : x 0 R < y < x 0 + R} M for all non-negative integers N, then the Taylor series (x 0 R, x 0 + R). f (n) (x 0 ) n! (x x 0 ) n of the function f converges to f(x) for all x
Remark: When you compute max{ f (N+) (y) : for all y between x 0 and x} you need to provide the tightest possible bound, else you will miss points. This bound may depend on x and x 0. In order to compute this bound, you will need to graph f (N+) and examine its monotonicity. There are two types of questions in Topic 0: Given a function f, a center x 0, a radius R, and a maximum power N of the Taylor polynomial, estimate the error f(x) p N (x) for all x in the interval (x 0 R, x 0 +R). Given a function f, a center x 0, a radius R, and a maximum allowable error, compute a power N of a Taylor polynomial such that the error f(x) p N (x) is at most equal to the maximum allowable error for all x in the interval (x 0 R, x 0 +R). Here without the use of calculator we will not be able to compute exactly the value of N. We can reduce the problem to solving an equation that we explain why it has a solution without solving it. 3
Topic : PARAMETRIC EQUATIONS OF PLANAR CURVES The questions that are encountered in this topic are the following: (a) Change the form of a curve, from parametric to Cartesian or backwards. (b) Compute the tangent line to a curve or examine the concavity of a curve. (c) Compute the length of a curve. (d) Compute the area between the curve and the x-axis, or between the curve and the y-axis. (e) Compute the area that is produced if you revolve a curve around the x-axis or around the y-axis. (a) To change the equation of a curve from parametric form to Cartesian form, you need to eliminate the time variable t, in order to produce an equation that contains only x and y. In order to make the reverse change, the following parametric equations may be useful and should be memorized. Commonly used curves: The graph of a function y = f(x): It can be parametrized by setting x = t. Thus we obtain x(t) = t and y(t) = f(t). A line passing from the point (a, b) and having slope m. It can be treated as a function and use the substitution x(t) = t. Another way to parametrize it is to use x(t) = t a and y(t) = mt + b. This parametrization has the property that (x(0), y(0)) = (a, b). A circle with center (x 0, y 0 ) and radius R has parametrization x(t) = x 0 + R cos t, y(t) = y 0 + R sin t, 0 t π. The Cartesian equation of the circle is (x a) + (y b) = R. The cycloid is the curve that is traced by a fixed point on a wheel of radius R when a wheel is rolling. The parametric equations of the cycloid are: x(t) = R(t sin t), y(t) = R( cos t). An ellipse with center (x 0, y 0 ), semi-major axis a and semi-minor axis b has Cartesian equation (x x 0) + (y y 0) =. Its parametric equations are x(t) = x a b 0 + a cos t, y(t) = y 0 + b sin t. (b) In order to compute the tangent line of a curve which is given in parametric form use the fact: dy dx = dy/dt dx/dt. In order to examine the concavity of a curve which is given in parametric form use the fact: d y dx = dy /dt dx/dt where y = dy dx. (c) The length of a curve which is given in parametric form is equal to s = b a x (t) + y (t) dt. (d) If y is above the x-axis, passes the vertical line test as t increases from t = a to t = b and the curve is traversed only once, then the area between the curve and the x-axis is given by Area = x(b) x(a) ydx = 4 b a y(t)x (t)dt.
Similarly, if x is on the right side of the y-axis, passes the horizontal line test as t increases from t = a to t = b and the curved is traversed only once, then the area between the curve and the y-axis is given by Area = y(b) y(a) xdy = b a x(t)y (t)dt. (e) If y is above the x-axis, and the curve is traversed only once as t increases from t = a to t = b, then the surface of the area by revolution of the curve around the x-axis, is given by the formula S = b a π y(t) x (t) + y (t) dt. Similarly, if x is on the right of the y-axis, and the curve is traversed only once as t increases from t = a to t = b, then the surface of the area by revolution of the curve around the y-axis, is given by S = b a π x(t) x (t) + y (t) dt. 5
Topic : POLAR COORDINATES In order to describe a point on the plane you can either give its Cartesian coordinates (x, y), or you can give its polar coordinates (r, θ). Every point on the plane has unique Cartesian coordinates but it has infinitely many polar coordinates. To compute one set of polar coordinates from the Cartesian coordinates, you set r = { arctan y x + y if (x, y) belongs in first or fourth quadrant, θ = x π + arctan y if (x, y) belongs in second or third quadrant. x Other sets of polar coordinates for the same point can be computed by adding any integer multiple of π to the above values of θ. Furthermore, more sets of polar coordinates for the same point can be obtained by adding (or subtracting) π from the above values of θ and simultaneously changing r to r = x + y. The reverse procedure of computing the (unique) (x, y) from (any set of ) (r, θ) is very simple: x = r cos θ, y = r sin θ. The questions that are encountered in this topic are the following: () Change an equation from Cartesian to polar or reverse. Assume that a curve is given in polar coordinates form r = f(θ) for some function f. Notice that x = r cos θ = f(θ) cos θ and y = r sin θ = f(θ) sin θ. () Compute its tangent line. (3) Compute the area of the fan-shaped region of the plane between the origin and the curve for a θ b. (4) Compute the length of the curve when the curved is traversed only once when θ increases from a to b. () Use the above definition of polar coordinates. () Use the formula dy dx = dy/dθ dx/dθ = f (θ) sin θ + f(θ) cos θ f (θ) cos θ f(θ) sin θ. (3) Use the formula b r b Area = a dθ = f(θ) a dθ. An immediate consequence of this formula is that if r = f (θ) and r = f (θ) are two curves with f (θ) f (θ) for all θ then the area of the fan-shaped region of the plane between these two curves for a θ b is equal to f (θ) f (θ) Area = dθ. a (4) The formula that we saw earlier simplifies to b s = f(θ) + f (θ) dθ. a b 6